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 first identify the anode (oxidation, lower \( E_{\text{red}} \)) and cathode (reduction, higher \( E_{\text{red}} \)). \( \text{Cr}^{3+} \) has \( E_{\text{red}} = -0.73 \, \text{V} \) (anode, so reverse: \( \text{Cr}
ightarrow \text{Cr}^{3+} + 3e^- \)), and \( \text{Cu}^{2+} \) has \( E_{\text{red}} = +0.34 \, \text{V} \) (cathode, \( \text{Cu}^{2+} + 2e^-
ightarrow \text{Cu} \)). Balance electrons: multiply Cr half - reaction by 2 (\( 2\text{Cr}
ightarrow 2\text{Cr}^{3+} + 6e^- \)) and Cu half - reaction by 3 (\( 3\text{Cu}^{2+} + 6e^-
ightarrow 3\text{Cu} \)). Add them: \( 3\text{Cu}^{2+}+2\text{Cr}
ightarrow 2\text{Cr}^{3+}+3\text{Cu} \), which matches option A.

Answer:

A