Question

solve the attachment???​

Answer 1

By Nernst Equation :

$\sf\:E_{cell}=E\degree_{cell}-\dfrac{0.09}{n}$

$\sf\implies\:E_{cell}=0.14-\dfrac{0.059}{2}$

$\sf\implies\:E_{cell}=0.14-\dfrac{0.056}{2}$

$\sf\implies\:E_{cell}=0.14-0.03[\log25+\log10]$

$\sf\implies\:E_{cell}=0.14-0.03[2\log5+\log10]$

$\sf\implies\:E_{cell}=0.14-0.03[2\times0.69-1]$

$</p><p>\sf\implies\:E_{cell}=0.14-0.03[1.38-1]$

$\sf\implies\:E_{cell}=0.14-0.03[0.38]$

$\sf\implies\:E_{cell}=0.14−0.0114$

$\sf\implies\:E_{cell}=0.14-0.0114$

$\sf\implies\:E_{cell}=0.1286V$

Therefore,the potential of the given cell is 0.1286V.

Answer 2

Answer:

$0.1286V$

Explanation:

The potential of the given cell is 0.1286V.