Consider the galvanic cell shown below (the contents of each half-cell are written beneath each compartment): 0.50 M Br₂ 0.20 M Cr³⁺ 0.10 M Br^- The standard reduction potentials are as follows: Cr³⁺(aq)+ 3e^- $\to$ Cr(s) $\varepsilon$ $\circ$ =-0.727 V Br₂(aq)+ 2e^- $\to$2Br^-(aq) $\varepsilon$ $\circ$ = +1.090 V What is $\varepsilon$ $\circ$ for this cell?

Question

Quizplus · Accepted Answer

To find the standard cell potential, we need to subtract the reduction potential of the anode from the reduction potential of the cathode:
E°cell = E°cathode - E°anode

The cathode (where reduction occurs) is the half-cell with Br^- ions:
E°cathode = +1.090 V

The anode (where oxidation occurs) is the half-cell with Cr^3+ ions:
E°anode = -0.727 V

Therefore,
E°cell = +1.090 V - (-0.727 V) = +1.817 V

Note that the signs of the reduction potentials are switched when we calculate the cell potential, because the reduction potential is a measure of the tendency to gain electrons (while in the cell, the anode loses electrons and the cathode gains electrons).

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