Question

Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.700 V when [Zn2+]= 0.26 M and PH2= 0.82 atm .

Answer 1

The pH of cathode compartment solution is 1.3995

Given:

cell emf at 298 K = 0.7 V

concentration $[Zn^{2+} ] = 0.26 M$

Pressure of $H_{2} = 0.82 atm$

n = 2

To Find:

The pH of the cathode compartment solution

Solution:

The reaction for the voltaic cell is as follows

$Zn(s)+2H^{+}(aq)$ ⇄ $H_{2}(g)+Zn^{2+}(aq)$

We shall use the Nernst equation to solve this.

$E=E^{o} - (\frac{0.0592}{n} )logQ,$

$Q_{eq}= \frac{P_{H_{2}}[Zn^{2+}]}{[H^{+}]^{2}}$

Lets calculate log $Q_{eq}$ first

$log Q_{eq} = log ( P_{H_{2}} [Zn^{2+}]) - log [H^{+}]^{2}$

= $log ( 0.82 * 0.26) - log [H^{+}]^{2}$

$= log 0.2132 - log [H^{+}]^{2}$

$= -0.671 - log [H^{+}]^{2}$

Now substitute this value in Nerst equation

$E=E^{o} - (\frac{0.0592}{n} ) ( -0.671 - log [H^{+}]^{2})$

The value of $E^{o}$=0.763V per standard tables.

The value of E = 0.7 V

The equation is now

$0.7=0.763 - (\frac{0.0592}{2} ) ( -0.671 - log [H^{+}]^{2})$

$\frac{2(0.763-0.7)}{0.0592} = ( -0.671 - log [H^{+}]^{2})$

$2.128 = -0.671 - log [H^{+}]^{2}$

$2.799=- log [H^{+}]^{2}$

$- 2*log [H^{+}] =2.799$

$- log [H^{+}] =\frac{2.799}{2}$

=1.3995

but $- log [H^{+}]$ is pH

So pH of cathode compartment solution is 1.3995

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Answer 2

Cathode compartment solution has a pH of $1.3995$.

Explanation:

Given information -

Cell emf at $298 k = 0.7 V$.

Concentration $[Zn^{2+} ] = 0.26M$.

Pressure of $H_{2} = 0.82 atm$.

To find -

The pH of the cathode compartment solution

Solution -

The voltaic cell's response is as follows:

$Zn(s) + 2H^{+} (aq)$ ⇄ $H_{2} (g) + Zn^{2+} (aq)$

To resolve this, we'll apply the Nernst equation.

$E = E^{0} - (\frac{0.0592}{n} ) logQ$

$Q_{eq} = \frac{P_{H_{2} }Zn^{2+} }{\{H^{+} ]^{2} } }$

First, calculate $Log Q_{1eq}$ :

$logQ_{eq} = log (P_{H_{2} } [Zn^{2+} ])- log[H^{+}]^{2}$

$= log ( 0.82 * 0.26) - log [H^{+} ]^{2}$

$= log 0.2132 - log [H^{+}] ^{2}$

$= -0.671 - log [Hx^{+} ] ^{2}$

Put this value in the Nerst equation now.

$E = E^{0} - (\frac{0.0592}{n} )(-0.671- log[H^{+} ]^{2}$

The value of $E^{0} = 0.763V$ per standard table.

The value of $E= 0.7 V$.

The equation now is as follows:

$0.7 = 0.763- (\frac{0.0592}{2} )(-0.671-log[H^{+} ]^{2}\\\frac{2(0.763-0.7)}{0.0592} = -0.671-log[H^{+} ]^{2}\\2.128 = -0.671-log[H^{+} ]^{2}\\2.799 = -log[H^{+} ]^{2}\\\\-2 * log[H^{+}]^{2} = 2.799\\ -log[H^{+} ]^{2} = \frac{2.799}{2}\\ = 1.3995$

But, $-log[H^{+}]$ is $pH$.

Therefore, the Cathode compartment solution has a pH of $1.3995$.

#SPJ2