Question

Calculate the emf of the cell in which the following reactions takes place:

Ni(s) + 2Ag+(0.002M) ------- Ni2+(0.160M) +2Ag(s)

Given that E0 (cell) = 1.05 V

Answer 1

Answer:

Explanation:

Let's get started..

Given cell reaction is

          Ni(s)+2Ag+(0.002M) ⇒Ni2+(0.160M)+2Ag(s)

And    E°cell=1.05V

i.e; [Ag+]=0.002M

     [Ni+2]=0.160M

We know that NERNST EQUATION is

          Ecell=E°cell-0.0591/n ㏒[ni2+]/[Ag+]²

Here n=2

Apply Nernst equation,

          Ecell=1.05-0.0591/2 ㏒(0.160)/(0.002)²

                   ⇒1.05-0.0295㏒(0.160)/(0.000004)

                   ⇒1.05-0.0295㏒(4x10^4)

                   ⇒1.05-0.0295{㏒4+4}

                   ⇒1.05-0.0295{0.6020+4}

                   ⇒1.05-0.0295{4.6020}

                   ⇒1.05-0.135

                   ⇒0.915V

Therefore,Ecell=0.915V

                             

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\green{Solution:}}}}}}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

Applying Nernst equation, we have:

$\sf E_{(cell)}\:=\:E_{(cell)}^{\Theta}- \dfrac{0.0591}{n}log \dfrac{[Ni^{2+}]}{[Ag^{+}]^2}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\sf E_{(cell)}\:=\:1.05- \dfrac{0.0591}{2}log \dfrac{(0.160)}{(0.002)^2}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\sf E_{(cell)}\:=\:1.05-log \dfrac{0.16}{0.000004}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\sf =\:1.05-0.02955\,log\,4\times 10^4$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\sf =\:1.05-0.02955(log\,10000+log\,4)$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\sf =\:1.05-0.02955(4+0.6021)$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\sf =\:0.914V$