The standard electrode potential for four half cells are as follows: Fe^2-(sq) +

Question

The standard electrode potential for four half cells are as follows: Fe^2-(sq) + 2c^- rightarrow Fe(s) E degree = -0.44 V Cd^2-(sq) + 2c^- rightarrow Cd(s) E degree = -0.403 V Cu^2+(aq) + 1 c^- rightarrow Cu^+(aq) E degree = +0.22 V MnO^-_4(sq) + 8 H^-(sq) + 5 c^- rightarrow Mn^2+(aq) + 4H_2O(I) E degree = +1.23 V Which chemical species is the most powerful reducing agent? Write a balanced cell reaction for the above reaction where Cd is oxidizied by MnO^-_4. Write the Nernst equation for a cell represented by the reaction in b). Calculate the cell voltage at 25 degree C for the cell in c) where the concentrations are as follows: [Cd^2+] = 0.100 M, [Mn^2+] = 0.0100 M, [MnO^-_4] = 0.100 M, and pH of solution is 7.00

Explanation / Answer

a)

most powerful reducing agent --> will oxidize itself readily, in order to favour reduction

This must be the species with the LEAST value in reuction optential

Choose Fe2+

b)

balance reaction with Cd, and MnO4

Cd+2(aq) + 2e- --> Cd(s)

MnO4-(aq) + 8H+(aq) + 5e- = Mn+2(aq) + 4H2O(l)

balance electrons

5Cd+2(aq) + 10e- --> 5Cd(s)

2MnO4-(aq) + 16H+(aq) + 10e- = 2Mn+2(aq) + 8H2O(l)

invert Cd equation (oxidation)

2MnO4-(aq) + 16H+(aq) + 10e- = 2Mn+2(aq) + 8H2O(l)

5Cd(s) -- > 5Cd+2(aq) + 10e-

add dall

5Cd(s) + 2MnO4-(aq) + 16H+(aq) + 10e- = 2Mn+2(aq) + 8H2O(l) + 5Cd+2(aq) + 10e-

cancel common terms

5Cd(s) + 2MnO4-(aq) + 16H+(aq)= 2Mn+2(aq) + 8H2O(l) + 5Cd+2(aq)

c)

Appl yNernst equation for B:

E = E° - 0.0592/n * log(Q)

E° = Ered - Eox = 1.23 - -0.403 = 1.633 V

E = 1.633 -  0.0592/10 * log(Q)

Q = [Mn+2][Cd+2] /([MnO4-][H+]^8)

E = 1.633 -  0.0592/10 * log([Mn+2][Cd+2] /([MnO4-][H+]^8))

d)

find V when

E = 1.633 -  0.0592/10 * log(0.01*0.1 / ((0.1)(10^-7)^8))

E = 1.31332 V

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