Question

A concentration cell consists of two Ag/Ag half-cells. In half-cell A, electrode ,A dips into 0.010 M AgNO3;
in half-cell B, electrode B dips into 4.0 X 10-4 M AgNO3. What is the cell potential at 298.15 K? Which
electrode has a positive charge?​

Answer 1

Answer:

Explanation:

Cell A

Answer 2

Given:

The concentration cell consists of two Ag/Ag half-cells.

In half-cell A, electrode , A dips into $\[0 \cdot 0100\,M\,AgN{O_3}\]$

In half-cell B, electrode B dips into $\[4 \times {10^{ - 4}}\,M\,AgN{O_3}\]$

To Find:

We have to find the cell potential and the electrode having a positive charge.

Solution:

$\[E_{cell}^0\]$ will be zero since the half-cell potentials are equal. $\[{E_{cell}}\]$ is calculated from the Nernst Equation with half-cell A having $\[A{g^ + }\]$ being reduced and plating out, and in half-cell, B Ag(s) will be oxidized to $\[A{g^ + }\]$.

     $\[\begin{array}{l}A{g^ + }\,\left( {aq,0 \cdot 0100\,M} \right)half - cell\,A \to A{g^ + }\,\left( {aq,4 \times {{10}^{ - 4}}\,M} \right)half - cell\,B\\\\{E_{cell}} = E_{cell}^0 - \frac{{0 \cdot 0592V}}{1}\log \frac{{{{\left[ {A{g^ + }} \right]}_{dilute}}}}{{{{\left[ {A{g^ + }} \right]}_{concentrated}}}}\\\\{E_{cell}} = 0 - 0 \cdot 0592\log 4 \times {10^{ - 2}}\\\\\,\,\,\,\,\,\,\,\,\, = 0 \cdot 0828V\end{array}\]$

∴The cell potential at 298·15K is 0·0828 V

Half-cell A is the cathode and has a positive electrode.