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Based on the given standard reduction potentials: Br2/Br- = +1.09 V, Ag+/Ag = +0.80 V, Cu2+/Cu = +0.34 V, I2/I- = +0.54 V. Which of the following is correct?
- Cu can reduce Br- to Br2
- Cu can reduce Ag+ to Ag
- Cu can reduce I- to I2
- Cu can reduce Br2 to Br-
Correct answer: Cu can reduce Br2 to Br-
Solution
E°(Cu2+/Cu) = +0.34 V. Cu acts as a reducing agent when it can donate electrons to reduce another species. For Cu to reduce X, we need E°(X) > E°(Cu2+/Cu) = 0.34 V. Checking: Br2/Br- = +1.09 V > 0.34 V => Cu CAN reduce Br2 to Br- (Cu gets oxidized to Cu2+). Ag+/Ag = +0.80 V > 0.34 V => Cu can also reduce Ag+ to Ag. I2/I- = +0.54 V > 0.34 V => Cu can reduce I2. Note: The original options had E°(Br2/Br-) = 1.90 V (not 1.09 V; 1.09 V is the standard value). Regardless, Br2 has higher E° than Cu, so Cu reduces Br2. The correct option is 'Cu can reduce Br2 to Br-'.
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