Free NEET Chemistry multiple-choice questions on Nernst Equation. Attempt each question and reveal the answer with a full explanation.
Which of the following expressions represents the Nernst equation for the half-reaction M n+ (aq) + ne - M(s) at 298 K ? E = E - 0.059 n 1 [M n+ ] E = E + 0.059 n [M n+ ] n E = E - 0.059 n [M n+ ] E = E + 0.059 [M n+ ] The equilibrium constant ( K ) for a cell reaction at 298 K where n=1 and E cell = 0.59 V is: 10 10 10 5 10 1 10 20 For the concentration cell Ag | Ag +(C 1) || Ag +(C 2) | Ag , the cell will be spontaneous only if: C 2 > C 1 C 1 > C 2 C 1 = C 2 C 1 = 0 For the cell reaction 2Fe 3+ (aq) + 2I -(aq) 2Fe 2+ (aq) + I 2(s) , E cell = 0.236 V at 298 K . The equilibrium constant ( K c ) of the cell reaction is: (Given 2.303RT/F = 0.059 ) 10 8 10 4 10 10 10 12 The standard reduction potential of Zn 2+ /Zn is -0.76 V . What is the potential of a Zn electrode in 0.01 M Zn 2+ solution at 25 C ? -0.819 V -0.701 V -0.76 V -0.85 V The equilibrium constant for the reaction Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) is 2 10 37 at 25 C . The standard EMF of the cell is approximately: 1.1 V 0.55 V 2.2 V 1.5 V What is the potential of a standard hydrogen electrode immersed in pure water at 298 K ? -0.414 V 0.000 V +0.414 V +0.828 V The potential of a silver electrode ( Ag/Ag + ) at 25 C in a 0.01 M AgNO 3 solution is (Given E Ag +/Ag = 0.80 V ): 0.682 V 0.859 V 0.741 V 0.918 V A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be 0.059 V 0.59 V 0.118 V 1.18 V The pressure of H2 required to make the potential of H2 -electrode zero in pure water at 298 K is :- 10 -14 atm 10 -12 atm 10 -10 atm 10 -4 atm For a cell involving one electron E cell = 0.59 V at 298 K, the equilibrium constant for the cell reaction is : [ Given that 2.303 RT F = 0.059 V at T = 298 K ] 1.0 10 30 1.0 10 2 1.0 10 5 1.0 10 10 The electrode potential of a hydrogen electrode depends on which of the following? All of these Concentration of H + ions Partial pressure of H 2 gas Temperature Find the emf of the cell in which the following reaction takes place at 298 K Ni(s) + 2Ag+ (0.001M) -> Ni 2+ (0.001M) + 2Ag(s) (Given that E cell = 1.05 V , 2.303 RT F = 0.059 at 298 K) 1.385 V 0.9615 V 1.05 V 1.0385 V Calculate emf of the half cell given below : Pt(s) | H2(g, 2 atm) | HCl(aq, 0.02 M) E H2/H+ = 0 V (Given: 2.303 RT F = 0.059 , 2 = 0.3010 ) -0.109 V 0.109 V 0.035 V -0.035 V A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10 -6 M hydrogen ions. The EMF of the cell is 0.118 V at 25 C . The concentration of hydrogen ions at the positive electrode is: 10 -4 M 10 -8 M 10 -2 M 10 -6 M For the reaction Fe(s) + Cu 2+ (0.01 M) Fe 2+ (0.1 M) + Cu(s) , the E cell is (Given E cell = 0.78 V ): 0.7505 V 0.8095 V 0.7800 V 0.7212 V Calculate the equilibrium constant K for the reaction Cu(s) + 2Ag +(aq) Cu 2+ (aq) + 2Ag(s) at 298 K (Given E cell = 0.46 V ): 3.92 10 15 2.0 10 10 1.5 10 12 4.0 10 7 For the cell reaction Cu(s) + 2Ag +(aq) Cu 2+ (aq) + 2Ag(s) , the equilibrium constant K c is 10 15.6 at 298 K . The E cell is approximately: 0.46 V 0.80 V 0.34 V 0.12 V The EMF of the concentration cell H 2(P 1) | H +(1M) || H +(1M) | H 2(P 2) at 298 K is given by: 0.0591 2 P 1 P 2 0.0591 2 P 2 P 1 0.0591 P 1 P 2 -0.0591 P 1 P 2 The standard reduction potential for the Cu 2+ /Cu couple is +0.34 V . What is the reduction potential at pH = 14 for this couple? (Given K sp of Cu(OH) 2 = 1.0 10 -19 ) -0.22 V +0.34 V -0.44 V +0.22 V The pressure of H 2 gas required to make the potential of a hydrogen electrode zero in 0.1 M HCl solution at 298 K is: 10 -2 atm 10 -1 atm 1 atm 10 -4 atm In a Cu/Cu 2+ half-cell, if ammonia ( NH 3 ) is added to the solution, the electrode potential ( E Cu 2+ /Cu ) will: Decrease Increase Remain the same First increase then decrease For the cell Zn | Zn 2+ || Cu 2+ | Cu , if the concentration of Zn 2+ and Cu 2+ ions are doubled, the cell potential will: Remain same Be doubled Be halved Become zero The temperature coefficient of the EMF of a cell is related to the entropy change ( S ) of the cell reaction by the expression: S = nF(dE/dT) S = nFE S = -nF(dE/dT) S = H / T The equilibrium constant of the reaction: Cu(s) + 2Ag+ (aq) -> Cu 2+ (aq) + 2Ag(s) ; E = 0.46 V at 298 K is 2.0 10 10 4.0 10 10 4.0 10 15 2.4 10 10 Which of the following conditions will increase the voltage of the cell Cu(s) + 2Ag +(aq) Cu 2+ (aq) + 2Ag(s) ? Increase in [Ag +] Increase in [Cu 2+ ] Increase in size of Ag electrode Decrease in size of Cu electrode In a concentration cell, the EMF is generated due to: Difference in concentration of electrolytes in the two half-cells Difference in nature of electrodes A chemical reaction occurring in the cell Difference in temperature between electrodes For a hydrogen electrode, the electrode potential at pH = 3 and P H 2 = 1 atm at 298 K is approximately: -0.177 V 0.00 V -0.059 V -0.236 V The equilibrium constant K for a cell reaction can be calculated from the standard cell potential E cell using the relation: K = n E cell 0.059 K = 0.059 n E cell K = -nFE cell K = E cell n 0.059 The electrode potential of a Hydrogen electrode at pH = 10 is: -0.59 V 0.00 V -0.059 V -1.18 V In the electrochemical cell Zn | Zn 2+ (0.01M) || Cu 2+ (1.0M) | Cu , the cell potential E cell will be: Greater than E cell Less than E cell Equal to E cell Zero