Question

what is the balanced chemical reaction for the cell with a potential of 1.07 v?
half-reaction | e_red (v)
cu²⁺ + 2e⁻ → cu | +0.34
cr³⁺ + 3e⁻ → cr | -0.73
a. 3cu²⁺ + 2cr → 2cr³⁺ + 3cu
b. 3cu²⁺ + 2cr³⁺ → 2cr + 3cu
c. cu + cr³⁺ → cu²⁺ + cr
d. 2cu + cr³⁺ → cr + 2cu²⁺
enter the answer choice letter.

Explanation:

To determine the balanced reaction, we analyze the reduction potentials. A positive \( E_{\text{red}} \) means the species is more likely to be reduced. \( \text{Cu}^{2+} \) has \( E_{\text{red}} = +0.34 \, \text{V} \) (more positive, so reduced), and \( \text{Cr}^{3+} \) has \( E_{\text{red}} = -0.73 \, \text{V} \) (so \( \text{Cr} \) will be oxidized, as oxidation is reverse of reduction: \( \text{Cr}
ightarrow \text{Cr}^{3+} + 3e^- \)).

For \( \text{Cu}^{2+} \) reduction: \( \text{Cu}^{2+} + 2e^-
ightarrow \text{Cu} \). To balance electrons, multiply Cu half - reaction by 3 (gives \( 6e^- \) gained) and Cr half - reaction by 2 (gives \( 6e^- \) lost). So:

• Oxidation (Cr): \( 2\text{Cr}

ightarrow 2\text{Cr}^{3+} + 6e^- \)

• Reduction (Cu): \( 3\text{Cu}^{2+} + 6e^-

ightarrow 3\text{Cu} \)

Adding these, we get \( 3\text{Cu}^{2+}+2\text{Cr}
ightarrow 2\text{Cr}^{3+}+3\text{Cu} \), which matches option A. Options B has \( \text{Cr}^{3+} \) being reduced (but its \( E_{\text{red}} \) is negative, so it should be oxidized), C and D have Cu being oxidized (but Cu's \( E_{\text{red}} \) is positive, so it should be reduced), so they are incorrect.

Answer:

A