In a galvanic cell, one half-cell consists of a lead strip dipped into a 1.00 M

Question

In a galvanic cell, one half-cell consists of a lead strip dipped into a 1.00 M solution of Pb(NO3)2. In the second half-cell, solid manganese is in contact with a 1.00 M solution of Mn(NO3)2. Pb is observed to plate out as the galvanic cell operates, and the initial cell voltage is measured to be 1.059 V at 25°C.

(a) Write balanced equations for the half-reactions at the anode and the cathode. Show electrons as e-. Use the smallest integer coefficients possible and the pull-down boxes to indicate states. If a box is not needed, leave it blank.

Half-reaction at anode (do not multiply by factor): (please indicate aq,liquid,solid,ext.)

Half-reaction at cathode (do not multiply by factor): (please indicate aq,liquid,solid,ext.)

(b) Calculate the standard reduction potential of a Mn2+|Mn half-cell. The standard reduction potential of the Pb2+|Pb electrode is -0.126 V.

Explanation / Answer

(a) Reaction occuing at cathode

Pb(2+)(aq) + 2e- -------------- Pb(s)

Reaction occuring at anode

Mn(s) ------------------ Mn(+2)(aq) + 2e-

Overall reaction

Pb(2+)(aq) + Mn(s) --------------- Mn(+2)(aq) + Pb(s)

b)

Eo(cell) = Eo(oxidation) + Eo(reduction)

1.059 = Eo(oxidation) - 0.126

Eo(oxidation) = 1.185

Hence the potential for  Mn2+|Mn cell will be -1.185V

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