In: Chemistry
What is the cell potential of the following cell at 25 oC?
Standard Reduction table
Cr / Cr3+ (0.053 M) // Ag1+ (0.52 M) / Ag
E=____________ V
Cr(s) -------------------------> Cr^3+ (aq) + 3e^- E0 = 0.74v
3Ag^+ (aq) + 3e^- ---------------> 3Ag(s) E0 = 0.80v
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Cr(s) + 3Ag^+ (aq) ------------------> Cr^3+ (aq) + 3Ag(s) E0 cell = 1.54v
n = 3
[Cr^3+] = 0.053M
[Ag^+] = 0.52M
Ecell = E0cell - 0.0592/n logQ
= 1.54- 0.0592/3 log[Cr^3+]/[Ag^+]^3
= 1.54 -0.0197log0.053/(0.52)^3
= 1.54 - 0.0197log0.376
= 1.54-0.0197*-0.4248
= 1.548v >>>>>answer