What is the cell potential of the following cell at 25 oC? Standard Reduction table Cr...

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

Solutions

Expert Solution

Cr(s) -------------------------> Cr^3+ (aq) + 3e^- E0 = 0.74v

3Ag^+ (aq) + 3e^- ---------------> 3Ag(s) E0 = 0.80v

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

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

queen_honey_blossom answered 1 year ago

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