Suppose you use Cu 2+ (aq)/Cu(s) E ° = 0.34 V and Sn 2+ (aq)/Sn(s) E ° = 0.14 V

Question

Suppose you use Cu2+(aq)/Cu(s) E° = 0.34 V and Sn2+(aq)/Sn(s) E° = 0.14 V half-cells as the basis of agalvanic electrochemical cell.
If the cell starts at standard conditions, and 0.400 amp of currentflows for 48.0 h, what are the concentrations of the dissolvedspecies at this point? Assume 1.00 L of solution.

These are the possible answers:

Suppose you use Cu2+(aq)/Cu(s) E° = 0.34 V and Sn2+(aq)/Sn(s) E° = 0.14 V half-cells as the basis of agalvanic electrochemical cell.
If the cell starts at standard conditions, and 0.400 amp of currentflows for 48.0 h, what are the concentrations of the dissolvedspecies at this point? Assume 1.00 L of solution.

These are the possible answers:

A. [Sn2+(aq)] =0.64 M, [Cu2+(aq)] = 1.36 M B. [Sn2+(aq)] =1.72 M, [Cu2+(aq)] = 0.28 M C. [Sn2+(aq)] =1.36 M, [Cu2+(aq)] = 0.64 M D. [Sn2+(aq)] =0.28 M, [Cu2+(aq)] = 1.72 M

Explanation / Answer

Standard conditions: 1 M Cu2+ and 1 M Sn 2+. Because it is 1 L,there is 1 mol Cu2+ and 1 mol Sn2+ Cu2+ + 2e- => Cu                       E = 0.34              Sn = > Sn2+ +2e-           E =0.14 Note 0.4 A = 0.4C/s. Also, 2 e- are reacted to form 1 mol Cu 48 hours *3600 s/hr * (0.4 C/s) *(1 mol e-/96485 C) * ( 1 mol Cu/2mol e-) = 0.358 mol Cu formed. Or 0.358 mol of Cu2+ consumed, 0.358 mol of Sn2+ formed. Thus, final Cu2+ = 1 - 0.358 = 0.64   => 0.64 MCu2+ final Sn2+ = 1 + 0.358 = 1.36     = > 1.36 MSn2+ The answer is C

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