Question

# Calculate $E_{\mathrm{cell}}^{\circ}$ for each balan ced redox reaction and determine if the reaction is spontaneous as written. a. $\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{Ag}(s) \longrightarrow 4 \mathrm{OH}^{-}(a q)+4 \mathrm{Ag}^{+}(a q)$ b. $\mathrm{Br}_{2}(l)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s)$ c. $\mathrm{PbO}_{2}(s)+4 \mathrm{H}^{+}(a q)+\mathrm{Sn}(s) \longrightarrow \mathrm{Pb}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Sn}^{2+}(a q)$

Solution

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Step 1
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PART A:

\begin{align*} \cancel{4\text{ e}^-}+\text{ O_2}(g)+2\text{ H_2O}(l) &\rightarrow 4\text{ OH^-}(aq) \text{ \space{ } }&E_{cat}^{\text{\textdegree}}&=0.40\text{ V}\\ 4\text{ Ag}(s)&\rightarrow 4\text{ Ag^{+}}(aq)+\cancel{4\text{ e}^-} \text{ \space{ } }&E_{an}^{\text{\textdegree}}&=0.80\text{ V}\\ \hline \text{ O_2}(g)+2\text{ H_2O}(l)+4\text{ Ag}(s)&\rightarrow 4\text{ OH^{-}}(aq)+4\text{ Ag^+}(aq)\text{ \space{ } }&E_{cell}^{\text{\textdegree}}&=? \end{align*}

\begin{align*} E_{cell}^{\text{\textdegree}} &= E_{cat}^{\text{\textdegree}}-E_{an}^{\text{\textdegree}}\\ &= 0.40\text{ V}-0.80\text{ V}\\ &= -0.40\text{ V} \end{align*}

Since $E_{cell}^{\text{\textdegree}}<0\text{ V}$, then this redox reaction is nonspontaneous in the forward direction.

1. Determine the half-reactions. Make sure to verify that they add up to the net reaction.

2. Look up standard reduction potentials for both half-reactions. Remember that oxidation takes place at the anode and reduction takes place at the cathode.

3. Calculate the standard cell potential, $E_{cell}^{\text{\textdegree}}$.

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