In a cell that utilises the reaction Zn(s) + 2H+(aq) - Zn2+(aq) + H2(g) addition of H2SO4 to cathod

Q: Why does adding $H_2SO_4$ increase $[H^+]$?

Sulfuric acid ($H_2SO_4$) is a strong acid, meaning it completely dissociates in water to release $H^+$ ions, thereby increasing the overall concentration of $H^+$ in the solution.

Q: Why doesn't the addition of acid affect the anode?

The acid is specifically added to the cathode compartment, where the reduction of $H^+$ to $H_2$ gas is taking place. The anode compartment contains $Zn^{2+}$ and is physically separated by a salt bridge.

Q: How does Le Chatelier's principle apply here?

Le Chatelier's principle states that a system at equilibrium will oppose any change. By adding $H^+$ (a reactant), the system shifts to the right (forward direction) to consume the excess $H^+$.

Q: Can we determine the change in $E_{cell}$ without the Nernst equation?

Yes! A shift in equilibrium to the right means the forward reaction becomes more spontaneous, which directly corresponds to an increase in the cell potential ($E_{cell}$).

Q: What would happen if we added a base instead?

Adding a base would neutralize the $H^+$ ions, decreasing their concentration. This would shift the equilibrium to the left and decrease the cell potential $E_{cell}$.

Question

In a cell that utilises the reaction
$Zn(s) + 2H^+(aq) \longrightarrow Zn^{2+}(aq) + H_2(g)$
addition of $H_2SO_4$ to cathode compartment will

Options:

(A) lower the E and shift equilibrium to the left

(B) lower the E and shift the equilibrium to the right

(C) increase the E and shift the equilibrium to the right

(D) increase the E and shift the equilibrium to the left

Sigma Percentile · Accepted Answer

### The Impact of Acid on Electrochemical Cells

Imagine you are operating an electrochemical cell, and you decide to play around with the chemical environment by adding a strong acid. How does the system react? Does the voltage drop, or does it surge? Let's dive into the fascinating interplay of chemical equilibrium and thermodynamics to find out.

### Analyzing the Setup

We are given a cell that operates on the following redox reaction:

$$Zn(s) + 2H^+(aq) \longrightarrow Zn^{2+}(aq) + H_2(g)$$

In this setup, Zinc acts as the anode (where oxidation occurs), and a Standard Hydrogen Electrode (SHE) acts as the cathode (where reduction occurs). The reaction quotient, $Q$, which tells us the ratio of products to reactants at any given moment, is written as:

$$Q = \frac{[Zn^{2+}]}{[H^+]^2}$$

Notice that pure solids and gases at standard pressure are omitted from the $Q$ expression, leaving us with just the aqueous ions.

### The Master Equation

To understand how the cell potential ($E_{cell}$) changes, we bring in our master tool—the Nernst equation:

$$E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log Q$$

Substituting our expression for $Q$, we get:

$$E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[H^+]^2}$$

We can also rewrite this by flipping the fraction inside the logarithm, which changes the sign of the term:

$$E_{cell} = E^\circ_{cell} + \frac{0.0591}{2} \log \frac{[H^+]^2}{[Zn^{2+}]}$$

### The Twist: Adding Sulfuric Acid

The problem states that we add $H_2SO_4$ to the cathode compartment. Sulfuric acid is a strong acid, meaning it completely dissociates in water to release $H^+$ ions.

**This causes a sudden and significant increase in the concentration of $H^+$ ions ($[H^+]$) in the solution.**

### Le Chatelier's Principle in Action

First, let's look at this from a purely chemical equilibrium perspective. According to Le Chatelier's principle, if a system at equilibrium experiences a change in concentration, it will shift to counteract that change.

By adding $H^+$ ions, we are essentially dumping more reactants into the system. To consume this excess reactant, the system will shift the equilibrium in the forward direction—**to the right**.

### The Mathematical Proof

Now, let's verify this intuition mathematically using our rearranged Nernst equation:

$$E_{cell} = E^\circ_{cell} + \frac{0.0591}{2} \log \frac{[H^+]^2}{[Zn^{2+}]}$$

Look closely at the logarithm term. The $[H^+]$ is in the numerator. As $[H^+]$ increases, the value of the fraction $\frac{[H^+]^2}{[Zn^{2+}]}$ increases. Consequently, the logarithm of this fraction also increases.

Since we are adding this larger positive value to $E^\circ_{cell}$, the overall cell potential **$E_{cell}$ must increase**.

### Final Verdict

Both our physical intuition (Le Chatelier's principle) and our mathematical proof (the Nernst equation) lead us to the exact same conclusion. Adding sulfuric acid increases the cell potential $E$ and shifts the equilibrium to the right. It is a beautiful harmony of thermodynamics and equilibrium!

Animated Solution for Chemistry - Electrochemistry: In a cell that utilises the reaction $Z n (s) + 2 H^{+} (a q) ⟶ Z n^{2 +} (a q) + H_{2} (g)$ addition of $H_{2} S O_{4}$ to cathode compartment will

Select Answer:

Visualized Solution

$C e l l R e a c t i o n$

$N er n s t E q u a t i o n$

$A dd i t i o n o f H_{2} S O_{4}$

$L e C ha t e l i e r^{'} s P r in c i pl e$

$E f f ec t o n E_{ce l l}$

$C o n c l u s i o n$

The Sigma Insight: Electrochemical Cells

The Impact of Acid on Electrochemical Cells

Analyzing the Setup

The Master Equation

The Twist

Le Chatelier's Principle in Action

The Mathematical Proof

Final Verdict

Similar Questions

Given the electrochemical cell: $E_{Cu^{2 +} ∣ Cu}^{\circ} = + 0.34 V$ $E_{Zn^{2 +} ∣ Zn}^{\circ} = - 0.76 V$ Identify the incorrect statement from the options below for the above cell.

For the galvanic cell, $Z n (s) + C u^{2 +} (0.02 M) ⟶ Z n^{2 +} (0.04 M) + C u (s)$ , $E_{C e l l} = \dots \times 1 0^{- 2} V$ . (Nearest integer) [Use $E_{C u / C u^{2 +}}^{\circ} = - 0.34 V$ , $E_{Z n / Z n^{2 +}}^{\circ} = + 0.76 V$ , $\frac{2.303 R T}{F} = 0.059 V$ ]

For the redox reaction $Z n (s) + C u^{2 +} (0.1 M) ⟶ Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{ce l l}^{\circ}$ is $1.10 V$ . $E_{ce l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$

If the standard electrode potential for a cell is $2 V$ at $300 K$ , the equilibrium constant ( $K$ ) for the reaction, $Zn (s) + Cu^{2 +} (a q) ⇌ Zn^{2 +} (a q) + Cu (s)$ at $300 K$ is approximately ( $R = 8 JK^{- 1} mol^{- 1}$ , $F = 96000 C mol^{- 1}$ )

The standard Gibbs energy for the given cell reaction in $kJ mol^{- 1}$ at $298 K$ is $Zn (s) + Cu^{2 +} (a q) ⟶ Zn^{2 +} (a q) + Cu (s), E^{\circ} = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C mol^{- 1}$ )

The reduction potential of hydrogen half-cell will be negative if

Consider the following $E^{\circ}$ values :\n $E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V$ \n $E_{S n^{2 +} / S n}^{\circ} = - 0.14 V$ \nUnder standard conditions, the potential for the reaction\n $S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)$ is

Animated Solution for Chemistry - Electrochemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a cell that utilises the reaction Zn(s)+2H+(aq)⟶Zn2+(aq)+H2​(g) addition of H2​SO4​ to cathode compartment will