Question

If the voltage of Zn-H+
cell is
0.542 V at 25 0 C, What is the
concentration of the [H+
]?
when : [Zn 2+] = 0.01 M and [H2
] =
1.0 M

Answer 1

$E^o_{cell} = E^o_{reduction} (H^+/H_2 ) - E^o_{reduction} (Zn^{2+} /Zn)$

$= 0.00 V - (-0.76 V) \\ = 0.76 V \\ E_{cell} = 0.542 \\$

the overall equation is

Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

so n = 2, and

So nernst equation :

$0.542 = 0.76 - (\frac{0.059}{2})logQ \\ -0.218 = -0.0295logQ \\ 7.3898 = log Q \\ Q= 10^7.3898 = 24535787.41 = 2.45 \times 10^7$

where Q = ([Zn⁺² ] [H₂] )/[H⁺]²

$2.45 \times 10^7 = \frac{0.01 \times 0.1}{[H^+]^2}$

Now we solve for [H⁺]

$[H+] = \sqrt{ 0.408 \times 10^{-10}}= 0.638 \times 10^{-5} \\ = 6.38 \times 10^{-6} \;M$