Question

What value of Q for the previous concentration cell would result in a voltage of 0.10 V? If the concentration of zinc ion at the cathode was 0.50 M, what was the concentration at the anode?

Short answer

The above reaction is spontaneous as \(\Delta {G^0} < 0\). The value of \(n = 6e\)

The value of \(Q = 1442.30\)

Step-by-step solution

Define the Standard potential cell

In electrochemistry, the Galvanic cellis a kind of electrochemical cell in which the current is produced using a redox reaction. Redox reactions involve oxidation as well as reduction. A galvanic cell consists of two half cells. In the one-half cell, oxidation occurs. This half-cell acts as the anode. In the other half cell, reduction occurs, This half-cell is termed as the cathode. These two half cells work together and constitute an electrochemical cell.

\({E^\circ }\)cell \( = {E^\circ }\) red ( cathode \() - {E^\circ }\) red ( anode )were,\({E^\circ }\) cell = standard emf of the cell.

\({E^\circ }\) red = standard reduction potential. And\({E^\circ }\) red ( cathode \() > {E^\circ }\) red( anode )

Determine the Balance equation

At anode half-cell.

\(Al(s) \to A{l^{3 + }} + 3{e^ - } \ldots \ldots \ldots .1\)

At cathode half-cell.

\(C{u^{2 + }} + 2{e^ - } \to Cu(s) \ldots \ldots ..2\)

Multiplying equation 1 by 2 and equation 2 by 3 and adding them.

\(\begin{aligned}{}2{\rm{Al}}(s) \to 2{\rm{Al}}{l^{3 + }} + 6{e^ - }\\3{\rm{C}}{{\rm{u}}^{2 + }} + 6e \to 3{\rm{Cu}}(s) - \ldots - \ldots - 3\end{aligned}\)

\((overall\)\({\rm{ reaction }}) \Rightarrow 2Al(s) + 3C{u^{2 + }} \to 2A{l^{3 + }} + 3Cu(s)\)

The value of \({\bf{n}} = {\bf{6}}{{\bf{e}}^ - }(\)answer)

Now calculating \(Q = {K_{eq}} = \frac{{{{\left( {A{l^{ + 3}}} \right)}^2}{{(Cu)}^3}}}{{{{(Al)}^2}{{\left( {{\rm{C}}{{\rm{u}}^{2 + }}} \right)}^3}}}\)

Here \(({\rm{Al}})\) solid and \(({\rm{Cu}})\) solid is taken as 1

\(\begin{aligned}{}Q = \frac{{{{\left( {A{l^{3 + }}} \right)}^2}}}{{{{\left( {C{u^{2 + }}} \right)}^3}}}\\Q = \frac{{{{(0.15)}^2}}}{{{{(0.025)}^3}}}\\Q = 1442.30({\rm{ answer }})\end{aligned}\)

To predict the spontaneity of the above reaction, we use the following equations,

\(\begin{aligned}{}\Delta {G^0} = \frac{{{\bf{0}}.{\bf{0591}}}}{{\bf{n}}}\log Q \ldots \ldots ..3\\\Delta {G^0} = - \frac{{0.0591}}{6}\log (1442.30)\\\Delta {G^0} = - 0.00985 \times 3.159\\\Delta {G^0} = - 0.031J\end{aligned}\)

The above reaction is spontaneous as \(\Delta {G^0} < 0\).