A voltaic cell utilizes the following reaction and operates at 298 K...?

Question

3Ce^4+ (aq) + Cr (s) -----> 3Ce^3+ (aq) + Cr^3+ (aq)

E^o (V) Values are:
Ce4+ (aq) + e- ----> Ce3+ (aq) = +1.61
Cr3+ (aq) + 3e- -----> Cr (s) = -.74

Questions:
1.) What is the emf of this cell when [Ce4+] = 1.8 M, [Ce3+] = 0.12 M , and [Cr3+] = 1.2×10^−2
2.) What is the emf of the cell when [Ce4+] = 1.1×10^−2M, [Ce3+] = 2.1M, and [Cr3+] = 1.6M
3Ce^4+ (aq) + Cr (s) -----> 3Ce^3+ (aq) + Cr^3+ (aq)

E^o (V) Values are:
Ce4+ (aq) + e- ----> Ce3+ (aq) = +1.61
Cr3+ (aq) + 3e- -----> Cr (s) = -.74

Questions:
1.) What is the emf of this cell when [Ce4+] = 1.8 M, [Ce3+] = 0.12 M , and [Cr3+] = 1.2×10^−2
2.) What is the emf of the cell when [Ce4+] = 1.1×10^−2M, [Ce3+] = 2.1M, and [Cr3+] = 1.6M

Explanation / Answer

E= 3E(Ce4+ - Ce3+) -E(Cr3+ - Cr)

= 3[Eo(Ce4+ - Ce) -0.059/1 log[Ce3+]/[Ce4+] ] - Eo(Cr3+-Cr)-0.059/3 log1/[Cr3+]

= 3[(1.61 -0.059log(0.12/1.8) ] - (-0.74 -0.0196 log(1/0.012)

= 4.948 + 0.777

= 5.725,

2) E = 3(1.61-0.059 log(2.1/0.011) ) -(-0.74 -0.0196 log(1/1.6))

= 4.42 + 0.736

= 5.156

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