Calculate E_cell for the following electrochemical cell at 25 °C. Pt(s) | Fe³⁺(aq,0.100 M) ,Fe²⁺(aq,0.040 M) || Cl^-(aq,0.50 M) | AgCl(s) | Ag(s)
The standard reduction potentials are as follows:
AgCl(s) + e^- → Ag(s) + Cl^-(aq)
E° = +0.222 V
Fe³⁺(aq) + e^- → Fe²⁺(aq)
E° = +0.771 V

Question

Quizplus · Accepted Answer

The cell potential, $E_{cell}$, can be calculated using the Nernst equation: $E_{cell} = E°_{cell} - \frac{0.0592}{n} \log Q$, where $E°_{cell}$ is the standard cell potential, $n$ is the number of moles of electrons transferred in the reaction, and $Q$ is the reaction quotient. Here, $E°_{cell} = E°_{cathode} - E°_{anode} = 0.222 V - 0.771 V = -0.549 V$. Since both half-reactions involve the transfer of 1 electron, $n = 1$. The reaction quotient, $Q$, for the iron half-reaction is $\frac{[Fe^{2+}]}{[Fe^{3+}]}$ = $\frac{0.040}{0.100}$. Plugging these values into the Nernst equation gives $E_{cell} = -0.549 V - \frac{0.0592}{1} \log \left(\frac{0.040}{0.100}\right)$, which calculates to approximately -0.555 V, taking into account the logarithm of 0.4 and the constants involved.

Calculate Ecell for the Following Electrochemical Cell at 25 °C

Calculate E_cell for the Following Electrochemical Cell at 25 °C