Question

Calculate EcellE_{\mathrm{cell}}^{\circ} for each balan ced redox reaction and determine if the reaction is spontaneous as written. a. O2(g)+2H2O(l)+4Ag(s)4OH(aq)+4Ag+(aq)\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. Br2(l)+2I(aq)2Br(aq)+I2(s)\mathrm{Br}_{2}(l)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s) c. PbO2(s)+4H+(aq)+Sn(s)Pb2+(aq)+2H2O(l)+Sn2+(aq)\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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Answered 2 years ago
Answered 2 years ago
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PART A:

4 e+ O2(g)+2 H2O(l)4 OH(aq)    Ecat°=0.40 V4 Ag(s)4 Ag+(aq)+4 e    Ean°=0.80 V O2(g)+2 H2O(l)+4 Ag(s)4 OH(aq)+4 Ag+(aq)    Ecell°=?\begin{align*} \cancel{4\text{ e}^-}+\text{ O$_2$}(g)+2\text{ H$_2$O}(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$_2$O}(l)+4\text{ Ag}(s)&\rightarrow 4\text{ OH$^{-}$}(aq)+4\text{ Ag$^+$}(aq)\text{ \space{ } }&E_{cell}^{\text{\textdegree}}&=? \end{align*}

Ecell°=Ecat°Ean°=0.40 V0.80 V=0.40 V\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 Ecell°<0 VE_{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, Ecell°E_{cell}^{\text{\textdegree}}.

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