Determine potential (in mV) of the cell: in which , and . Given: ; ; — Electrochemistry Chemistry Question
Question
Determine potential (in mV) of the cell: $Pt \| Fe^{2+}, Fe^{3+} \|\| Cr_2O_7^{2-}, Cr^{3+}, H^+ \| Pt$ in which $[Fe^{2+}] = 0.75 \text{ M}, [Fe^{3+}] = 0.75 \text{ M}$, $[Cr_2O_7^{2-}] = 2 \text{ M}, [Cr^{3+}] = 4 \text{ M}$ and $[H^+] = 1 \text{ M}$. Given: $Fe^{3+} + e^- \rightarrow Fe^{2+}; E^\circ = 0.77 \text{ V}$; $14 H^+ + 6e^- + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O; E^\circ = 1.35 \text{ V}$; $2.303RT/F = 0.06, \log 2 = 0.3, \log 3 = 0.48$
💡 Solution & Explanation
The cell reaction is $6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$, with $n = 6$. $E^\circ_{cell} = 1.35 - 0.77 = 0.58$ V. Reaction quotient $Q = ([Fe^{3+}]^6 [Cr^{3+}]^2) / ([Fe^{2+}]^6 [Cr_2O_7^{2-}][H^+]^{14})$. Since $[Fe^{3+}] = [Fe^{2+}]$, they cancel out. $Q = 4^2 / (2 \times 1^{14}) = 16 / 2 = 8$. $E_{cell} = E^\circ_{cell} - (0.06/6) \log Q = 0.58 - 0.01 \log 8 = 0.58 - 0.01(0.9) = 0.58 - 0.009 = 0.571$ V = 571 mV. Therefore, correct answer is 0571.