Let's break down the problem step-by-step:
### i) Draw the galvanic cell
To draw the galvanic cell, we can provide a detailed description since a physical drawing isn't feasible in this format. A galvanic cell involves two half-cells connected by a salt bridge or a porous membrane. In this case, the galvanic cell consists of a zinc electrode and a standard hydrogen electrode (SHE).
Description of the galvanic cell:
- Anode: Zinc electrode (Zn)
- Anode reaction (oxidation): Zn → Zn²⁺ + 2e⁻
- Cathode: Standard Hydrogen Electrode (SHE)
- Cathode reaction (reduction): 2H⁺ + 2e⁻ → H₂
- Salt bridge: To complete the circuit
The overall cell is described as:
```
Zn electrode | Zn²⁺ (solution) || H⁺ (solution) | Standard Hydrogen Electrode (SHE)
```
### ii) Write down the complete cell reaction
To find the overall cell reaction, we combine the oxidation reaction at the anode and the reduction reaction at the cathode.
Anode (oxidation) reaction:
[tex]\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^- \][/tex]
Cathode (reduction) reaction:
[tex]\[ 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \][/tex]
Adding these two half-reactions together, the electrons cancel out:
[tex]\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \][/tex]
So, the complete cell reaction is:
[tex]\[ \text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \][/tex]
### iii) Calculate the emf of the cell
To calculate the electromotive force (emf) of the cell, we use the standard electrode potentials for the two half-cells.
Given standard reduction potentials:
- For the zinc electrode ([tex]\( \text{Zn}^{2+}/\text{Zn} \)[/tex]): [tex]\( E^\circ = -0.76 \text{ V} \)[/tex]
- For the standard hydrogen electrode ([tex]\(\text{SHE}\)[/tex]): [tex]\( E^\circ = 0.00 \text{ V} \)[/tex]
The emf of the cell is calculated using the formula:
[tex]\[ \text{EMF of the cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \][/tex]
Here:
- The cathode is the SHE: [tex]\( E^\circ_{\text{cathode}} = 0.00 \text{ V} \)[/tex]
- The anode is the Zn: [tex]\( E^\circ_{\text{anode}} = -0.76 \text{ V} \)[/tex]
Thus:
[tex]\[ \text{EMF of the cell} = 0.00 \text{ V} - (-0.76 \text{ V}) = 0.76 \text{ V} \][/tex]
Therefore, the emf of the cell is 0.76 V.
