Question

Represent the cell in which the following reaction takes place
Mes) + 2Ag (0.0001 M) → Mg**(0.130M) + 2Ag(s)
Calculate its Ecelif E
= 3.17 V.​

Answer 1

Given:

A reaction cell has been provided involving Silver and Magnesium.

To find:

• Represent the cell reaction

• EMF of cell at that concentration

Calculation:

Cell reaction:

$\sf{Mg(s) + 2Ag^{+}\rightarrow Mg^{+2} + 2Ag(s)}$

Let EMF of cell be denoted be $E_{cell}$

$\therefore \: E_{cell} = {E}^{ \: 0} _{cell} - \dfrac{0.059}{n} \bigg \{ log(Q) \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ log(Q) \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ log\frac{{Mg}^{ + 2} }{ {({Ag}^{ + }) }^{2} } \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ log\frac{0.13 }{ {(0.0001) }^{2} } \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ log\frac{0.13 }{ {( {10}^{ - 4} ) }^{2} } \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ log\frac{0.13 }{ {( 10}^{ - 8} )} \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ log(0.13 \times {10}^{8}) \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ log(13 \times {10}^{6}) \bigg \}$

$= > \: E_{cell} = 3.17 - \dfrac{0.059}{(2)} \bigg \{ 7.11\bigg \}$

$= > \: E_{cell} = 3.17 - 0.209$

$= > \: E_{cell} = 2.96 \: volt$

So, final answer is:

$\boxed{ \sf{ \: E_{cell} = 2.96 \: volt}}$