In the following electrochemical cell, what is the cathode half reaction? (aq)||

Question

In the following electrochemical cell, what is the cathode half reaction? (aq)||Fe^3+ (aq), Fe^2+ (aq)|Pt(s) a. Fe^2+ (aq) + e^- rightarrow Fe^3+ (aq) b. Fe^3+ (aq) +e^- rightarrow Fe^2+ (aq) c. Mn(s) rightarrow Mn^2+ (aq) + 2e^- d. Fe^2+ (aq) + Pt (s) rightarrow Fe^3+ (aq) + e^- e. Mn^2+ (aq) rightarrow Mn(s) + 2e^- Given: Pb^2+ (aq) + 2e^- reversiblearrow Pb(s); E degree = -0.13 V 2H^- (aq) + 2e^- reversiblearrow H_2(g); E degree = .00V NO_3^- (aq) + 4H^+ (aq) + 3e^- reversiblearrow NO (g) + 2H_2O (l); E degree = 0.96 V O_2 (g) + 4H^+ (aq) + 4e^- reversiblearrow 2H_2O (l); R degree = 1.23 V PbO_2 (s) + SO_4^2- (aq) + 4H^+ (aq) + 2e^- reversiblearrow PbSO_4(s) + 2H_2O (l); E degree = 1.69 V Under standard-state conditions, which of the following is the best oxidizing agent? A. O_2 b. PbSO_4 c. NO_3^- d. Pb^2+ e. PbO_2 In any electrochemical cell, the anode is always _______. a. the positive electrode b. the negative electrode c. the electrode at which some species gains electrons d. the electrode at which some species loses electrons e. the electrode at which reduction occurs The cell potential of an electrochemical cell with the cell reaction below is 1.607 V. 2Al (s) + 3Zn^2+ (aq) rightarrow 3Zn(s) + 2Al^3+ (aq) What is the maximum electrical work obtainable from this cell when 0.50g of Al is consumed a. -6.3 times 10^6 J b. -1.7 times 10^4 J Which reaction would voltaic cell? a. PbSO_4 (s) + 2e^- rightarrow Pb(s) + SO_4^2- (aq) b. 2H_2O(l) rightarrow O_2(g) + 4H^+ (aq) + 4e^- c. PbSO_4(s) rightarrow Pb^2+ (aq) + SO_4^2- (aq) d. 2H_2O(l) rightarrow 2H_2(g) + O_2(g) e. 2H_2O (l) + 2e^- rightarrow H_2(g) + 2OH^- (aq)

Explanation / Answer

Answer:

14. In this cell two types of reaction occurs

Oxidation(Anode): Mn ------------> Mn2+ + 2e-

Reduction(Cathode): Fe3+ + e- ------------> Fe2+

15. PbO2 having the largest positive value of electrode potential, is the strongest oxidizing agent.

Note: As per chegg guidelines solve first question.

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