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The free energy change for the following reaction at $25C,$ when $[C{u}^{2+}]=1\mathrm{.}16M$ and $[P{b}^{2+}]=0\mathrm{.}00284M,$ is $-104kJ$ :

$C{u}^{2+}(1\mathrm{.}16M)+Pb(s)Cu(s)+P{b}^{2+}(0\mathrm{.}00284M),G=-104kJ$

What is the cell potential for the reaction as written under these conditions?

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

Would this reaction be spontaneous in the forward or the reverse direction?

forward direction

reverse direction