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A voltaic cell is constructed from an Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s) half-cell.The initial concentration of...

A voltaic cell is constructed from an Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s) half-cell.The initial concentration of Ni2+(aq) in the Ni2+−Ni half-cell is [Ni2+]= 1.80×10−2 M . The initial cell voltage is +1.13 V .

1. By using data in Table 20.1 in the textbook, calculate the standard emf of this voltaic cell.

Solutions

Expert Solution

Ni(s) --------------> Ni^2+ + 2e^-          E0 = 0.23v

2Ag^+(aq)+ 2e^- ----------> 2Ag(s)      E0 = 0.80v

-----------------------------------------------------------------------

Ni(s) + 2Ag^+(aq) ---------------> 2Ag(s) + Ni^2+ (aq) E0cell = 1.03v

n= 2

Ecell = E0cell -0.0592/n logQ

Ecell     = E0cell- 0.0592/2 log[Ni^2+]/[Ag^+]^2

1.13      = 1.03 -0.0296log0.018/[Ag^+]^2

1.13-1.03   = -0.0296log0.018/[Ag^+]^2

-0.1/0.0296   = log0.018/[Ag^+]^2

log0.018/[Ag^+]^2   = -3.3783

0.018/[Ag^+]^2      = 10^-3.3783

0.018/[Ag^+]^2     = 0.0001485

[Ag^+]^2             = 0.018/0.0001485

[Ag^+]^2               = 121.2

[Ag^+]                   = 11M >>>>>answer

queen_honey_blossom answered 3 years ago
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