$Fe(s)|F{e}^{2+}(aq)||NO(g)|N{O}_3^{-}(aq),H^{+}(aq)|Pt(s)|$

$Fe(s)+2N{O}_3^{-}(aq)+4H^{+}(aq)F{e}^{2+}(aq)+NO(g)+2H_{2}O(l)$

$3Fe(s)+2N{O}_3^{-}(aq)+8H^{+}(aq)3F{e}^{2+}(aq)+2NO(g)+4H_{2}O(l)$

$3Fe(s)+N{O}_3^{-}(aq)+8H^{+}(aq)3F{e}^{2+}(aq)+2NO(g)+H_{2}O(l)$

$Fe(s)+N{O}_3^{-}(aq)+4H^{+}(aq)F{e}^{2+}(aq)+NO(g)+2H_{2}O(l)$

Previous Answers

Correct

Separate the line notation into two half$-$cell reactions by considering that the oxidation half$-$reaction components are on the left and the reduction half$-$reaction components are on the right. Balance each half$-$reaction with respect to charge by adding electrons:

Oxidation: $,Fe(s)F{e}^{2+}(aq)+2e^{-}$

Reduction: $,N{O}_3^{-}(aq)+4H^{+}(aq)+3e^{-}NO(g)+2H_{2}O(l)$

Make the number of electrons in both half$-$reactions equal by multiplying the oxidation half$-$reaction by $3,$ and the reduction half$-$reaction by $2$ :

$3Fe(s),3F{e}^{2+}(aq)+6e^{-}$

$2N{O}_3^{-}(aq)+8H^{+}(aq)+6e^{-},2NO(g)+4H_{2}O(l)$

Add the two half$-$reactions together and cancel the electrons to get the overall balanced reaction:

$3Fe(s)3F{e}^{2+}(aq)+6e^{\text{'}}$

$2N{O}_3^{-}(aq)+8H^{+}(aq)+6e^{-}2NO(g)+4H_{2}O(l)$

$3Fe(s)+2N{O}_3^{-}(aq)+8H^{+}(aq)3F{e}^{2+}(aq)+2NO(g)+4H_{2}O(l)$

Part B

Calculate $E_{\text{c}\text{e}\text{l}\text{l}}$ using the tabulated standard electrode potentials at $25C.$

Express your answer in volts to three significant figures.

$E_{\text{c}\text{e}\text{l}\text{l}}=$

V