Question

Calculate E cell and K for electrochemical cell : Mg| MgSO4 (.2M) || CuSO4(. 1M) | Cu at 27°C​

Answer 1

$E_{cell}$ and K are 2.71V and $1.1\times 10^{91}$

Explanation:

$Mg/Mg^{2+}(0.2M)//Cu^{2+}(0.1M)/Cu$

Mg behave as anode and copper behave as cathode

First we have to calculate the standard electrode potential of the cell.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{\text {anodic ion}}{\text {cathodic ion}}$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = $27^oC=273+27=300K$

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential of the cell = ?

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=(E^o_{cathode}-E^0{anode})-\frac{2.303\times (8.314)\times (300)}{2\times 96500}\log \frac{0.2}{0.1}$

$E^o_{cell}=0.34-(-2.38)=2.72V$

Using Nernest equation :

$E_{cell}=2.72-0.00895=2.71V$

Relation between standard Gibbs free energy and emf follows:

$Delta G^o=-nFE$

$Delta G^o=-2\times 96500\times 2.71=-523230$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^0=-RT\ln K$

$-523230=-8.314\times 300\ln K$

$K=1.1\times 10^{91}$

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