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

3 Ce4+(aq) + Cr(s) 3 Ce3+(aq) + Cr3+(aq)

1) What is the emf of this cell under standard conditions?


2) What is the emf of this cell when [Ce4+] = 1.5 M, [Ce3+] = 0.015 M, and [Cr3+] = 0.010 M?

3) What is the emf of the cell when [Ce4+] = 0.52 M, [Ce3+] = 0.83 M, and [Cr3+] = 1.3 M?

2007-05-07 15:17:43 · 1 answers · asked by Destiny E 2 in Science & Mathematics Chemistry

1 answers

K = [Cr3+][Ce3+]^3/[Ce4+]^3

And use the Nernst eqn.

2007-05-07 22:48:51 · answer #1 · answered by ag_iitkgp 7 · 1 8

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