Question

Consider a galvanic cell using solid $Cu$ and $Fe$ metals with their corresponding solutions.
What is the $E_{cell}^{\circ }$?

Standard Potential (V)	Reduction Half-Reaction
$2.87$	$F_2\left(g\right)+2e^{-}\to 2F^{-}\left(aq\right)$
$1.51$	$Mn{O}_4^{-}\left(aq\right)+8H^{+}\left(aq\right)+5e^{-}\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)$
$1.36$	$C{l}_2\left(aq\right)+3e^{-}\to 2C{l}^{-}\left(aq\right)$
$1.33$	$C{r}_2{O}_7^{2-}\left(aq\right)+14H^{+}\left(aq\right)+6e^{-}\to 2C{r}^{3+}\left(aq\right)+7H_{2}O\left(l\right)$
$1.23$	$O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}\to 2H_{2}O\left(l\right)$
$1.06$	$B{r}_2\left(l\right)+2e^{-}\to 2B{r}^{-}\left(aq\right)$
$0.96$	$N{O}_3^{-}\left(aq\right)+4H^{+}\left(aq\right)+3e^{-}\to NO\left(g\right)+H_{2}O\left(l\right)$
$0.80$	$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$$
$0.77$	$F{e}^{3+}\left(aq\right)+e^{-}\to F{e}^{2+}\left(aq\right)$
$0.68$	$O_2\left(g\right)+2H^{+}\left(aq\right)+2e^{-}\to H_2{O}_2\left(aq\right)$
$0.59$	$Mn{O}_4^{-}\left(aq\right)+2H_{2}O\left(l\right)+3e^{-}\to Mn{O}_2\left(s\right)+4O{H}^{-}\left(aq\right)$
$0.54$	$I_2\left(s\right)+2e^{-}\to 2I^{-}\left(aq\right)$
$0.40$	$O_2\left(g\right)+2H_{2}O\left(l\right)+4e^{-}\to 4O{H}^{-}\left(aq\right)$
$0.34$	$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$
$0$	$2H^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right)$
$-0.28$	$N{i}^{2+}\left(aq\right)+2e^{-}\to Ni\left(s\right)$
$-0.44$	$F{e}^{2+}\left(aq\right)+2e^{-}\to Fe\left(s\right)$
$-0.76$	$Z{n}^{2+}\left(aq\right)+2e^{-}\to Zn\left(s\right)$
$-0.83$	$2H_{2}O\left(l\right)+2e^{-}\to H_2\left(g\right)+2O{H}^{-}\left(aq\right)$
$1.66$	$A{l}^{3+}\left(aq\right)+3e^{-}\to Al\left(s\right)$
$-2.71$	$N{a}^{+}\left(aq\right)+e^{-}\to Na\left(s\right)$
$-3.05$	$L{i}^{+}\left(aq\right)+e^{-}\to Li\left(s\right)$

• A. $-0.78V$
• B. $0.10V$
• C. $0.78V$
• D. $-0.10V$

Answer 1

The correct option is C $0.78V$

$E^0$ for Cu = 0.34 V

$E^0$ for Fe = -0.44V

so, $E_{cell}^0=0.34-(-0.44)=0.78V$