Question

Questions

=> at 298k
pt(s) | H₂(g , 1 bar)| H⁺(aq , 1M ) || m⁴⁺(aq) | m²⁺(aq)

E꜀ₑₗₗ= 0.092v , when [ m²⁺ ( aq ) / m⁴⁺( aq ) ] = 10ˣ

Find the value of x

where E⁰m⁴⁺/m²⁺ = 0.151v
​

Answer 1

Given:

• pH₂ = 1 atm
• Concentration of H⁺ = 1 M
• E꜀ₑₗₗ= 0.092 V
• M²⁺ ( aq ) / M⁴⁺( aq )  = 10ˣ
• E⁰m⁴⁺/m²⁺ = 0.151v

Solution:

⇒ Pt(s) | H₂(g , 1 bar)| H⁺(aq , 1M ) || M⁴⁺(aq) | M²⁺(aq)

Here, LHS represents the reaction taking place at the anode and RHS represents the reaction taking place at the cathode.

Trick to remember:  ABC ( Anode-bridge-cathode)

Reaction at anode: H₂ → 2H⁺ + 2e⁻

Reaction at cathode: M⁴⁺+ 2e⁻ → M²⁺

Net reaction: H₂ +  M⁴⁺ →  M²⁺ +  2H⁺

The standard electrode potential of the cell is given by,

E°꜀ₑₗₗ = E꜀ - Eₐ

here,

• E꜀  = standard reduction potential at the cathode
• Eₐ = standard reduction potential at the anode

⇒ E°꜀ₑₗₗ = 0.151 -0

⇒ E°꜀ₑₗₗ = 0.151 V

Now,

By using the Nernst equation,

$\blacksquare\:$ E꜀ₑₗₗ=   E°꜀ₑₗₗ  -  RT/nF (ln Q)

here,

• R = universal gas constant = 8.314 J/mol K
• T = temperature
• n = no of moles of electrons exchanged= 2
• F = faraday's constant = 96500 C
• Q = Reaction quotient

RT/F = 8.314 x 298 / 96500 = 0.0257

Now,

E꜀ₑₗₗ=   E°꜀ₑₗₗ  -  0.0257/n  (ln Q)

E꜀ₑₗₗ=   E°꜀ₑₗₗ  -  0.0257 x2.303 /n ( log Q)

E꜀ₑₗₗ=   E°꜀ₑₗₗ  - 0.059 / n (log Q)

$\to \sf{E_{cell} = E^{\circ}_{cell} - \dfrac{0.059}{n}. log \dfrac{[H^{+}]^2[M^{2+}]}{pH_2 .[M^{4+}]} }$

Now let's substitute the given values in the above equation,

$\to \sf{0.092 = 0.151 - \dfrac{0.0592}{2}. log\dfrac{(1)^2 \times 10^x}{1 }}$

$\to \sf{0.059 = \dfrac{0.0592}{2}. log 10^x}$

$\to \sf{2 = log 10^x}$

$\to \sf{2 = x. log 10}$

$\to \boxed{\sf{x = 2 }}$

The value of x is 2.

Answer 2

Given :

• Pressure of Hydrogen = 1 atm

• Concentration of Hydrogen ion = 1 M

• $\sf{E_{cell}=0.092\:V}$

• $\sf{\mathsf{M}^{2+}(\mathsf{aq}) / \mathsf{M}^{4+}(\mathsf{aq})=10^{\mathsf{x}}}$

• $\sf{\mathsf{E}^{0} \mathsf{~m}^{4+} / \mathsf{m}^{2+}=0.151 \mathsf{~V}}$

Answer :

The cell is represented as,

$\sf{Pt(\mathsf{s})\left|\mathsf{H}_{2}(\mathsf{~g}, 1 \mathsf{bar})\right| \mathsf{H}^{+}(\mathsf{aq}, 1 \mathsf{M}) \| \mathsf{M}^{4+}(\mathsf{aq}) \mid \mathsf{M}^{2+}(\mathsf{aq})}$

Anode Half Cell Reaction : $\sf{\mathsf{H}_{2} \longrightarrow 2 \mathsf{H}^{+}+2 \mathsf{e}^{-}}$

Cathode Half Cell Reaction : $\sf{\mathsf{M}^{4+}+2 \mathsf{e}^{-} \longrightarrow \mathsf{M}^{2+}}$

​Overall Reaction is $\bf{\mathbf{H}_{2}+\mathbf{M}^{4+} \longrightarrow \mathbf{M}^{2+}+2 \mathbf{H}^{+}}$

By Using Nernst Equation

$\sf{\displaystyle E_{cell}=E_{cell}^{0}-\frac{0.059}{\mathsf{n}} \log \frac{\left[\mathsf{H}^{+}\right]^{2}\left[\mathsf{M}^{2+}\right]}{\mathsf{pH}_{2} \cdot\left[\mathsf{M}^{4+}\right]}}$

Substitute the known Values in the above Equation,

$\sf{\displaystyle 0.092=0.151-\frac{0.0592}{2} \log \frac{1^{2} \times 10^{\mathsf{x}}}{1}}$

$\to\sf{\displaystyle 0.092=0.151-\frac{0.0592}{2} \log \frac{1 \times 10^{\mathsf{x}}}{1}}$

$\to\sf{\displaystyle 0.092=0.151-\frac{0.0592}{2} \log \frac{10^{\mathsf{x}}}{1}}$

$\to\sf{\displaystyle 0.092=0.151-\frac{0.0592}{2} \log 10^{\mathsf{x}}}$

$\to\sf{\displaystyle 0.092-0.151=0.151-\frac{0.0592}{2} \log 10^{\mathsf{x}}-0.151}$

$\to\sf{\displaystyle 0.092-0.151=-\frac{0.0592}{2} \log 10^{\mathsf{x}}}$

$\to\sf{\displaystyle -0.059=-\frac{0.0592}{2} \log 10^{\mathsf{x}}}$

$\to\sf{\displaystyle -0.059\times -1=\left(-\frac{0.0592}{2} \log 10^{\mathsf{x}}\right) \times -1}$

$\to\sf{\displaystyle 0.059=-\frac{0.0592}{2} \log 10^{\mathsf{x}}}$

$\to\sf{2=\log 10^{x}}$

$\to\sf{2=x \log 10}$

$\to\sf{x=2}$