Copper is oxidized by nitric acid. If this property were used in an electrochemical cell, what would the standard cell potential be? The relevant reduction reactions and standard reduction potentials are given below. $$\begin{array} { l c }
+ 0.34 \mathrm {~V} & \mathrm { Cu } ^ { 2 + } + 2 \mathrm { e } ^ { - } \rightarrow \mathrm { Cu } \
+ 0.96 \mathrm {~V} & \mathrm { NO } _ { 3 } { } ^ { - } + 4 \mathrm { H } ^ { + } + 3 \mathrm { e } ^ { - } \rightarrow \mathrm { NO } + 2 \mathrm { H } _ { 2 } \mathrm { O }
\end{array}$$

Question

Copper is oxidized by nitric acid. If this property were used in an electrochemical cell, what would the standard cell potential be? The relevant reduction reactions and standard reduction potentials are given below. $$\begin{array} { l c } 
+ 0.34 \mathrm {~V} & \mathrm { Cu } ^ { 2 + } + 2 \mathrm { e } ^ { - } \rightarrow \mathrm { Cu } \
+ 0.96 \mathrm {~V} & \mathrm { NO } _ { 3 } { } ^ { - } + 4 \mathrm { H } ^ { + } + 3 \mathrm { e } ^ { - } \rightarrow \mathrm { NO } + 2 \mathrm { H } _ { 2 } \mathrm { O }
\end{array}$$

Quizplus · Accepted Answer

The standard cell potential is calculated by subtracting the standard reduction potential of the oxidation half-reaction from the reduction half-reaction: $ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} = 0.96\, \text{V} - 0.34\, \text{V} = 0.62\, \text{V} $.

Copper Is Oxidized by Nitric Acid A) -062 V B) +0

Copper Is Oxidized by Nitric Acid A) -062 V
B) +0