<u>Given:</u>
Concentration of Cr2+ = 0.892 M
Concentration of Fe2+ = 0.0150 M
<u>To determine:</u>
The cell potential, Ecell
<u>Explanation:</u>
The half cell reactions for the given cell are:
Anode: Oxidation
Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V
Cathode: Reduction
Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V
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Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)
E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V
The cell potential can be deduced from the Nernst equation as follows:
Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]
Here, n = number of electrons = 2
Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V
Ans: The cell potential is 0.418 V
