Consider the half-cell equations and their half-cell potentials in the table below - VCE - SSCE Chemistry - Question 28 - 2023 - Paper 1
Question 28
Consider the half-cell equations and their half-cell potentials in the table below.
| Half-cell equations | Standard electrode potential (E°) i... show full transcript
Worked Solution & Example Answer:Consider the half-cell equations and their half-cell potentials in the table below - VCE - SSCE Chemistry - Question 28 - 2023 - Paper 1
Step 1
Determine the E° values for each half-cell
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
From the provided table,
The reduction potential for Mn²⁺(aq) + e⁻ ⇌ Mn(s) is +1.56 V.
The reduction potential for SO₃²⁻(aq) + 3H₂O(l) ⇌ SO₄²⁻(aq) + 6OH⁻(aq) is -0.57 V.
Using these values, we can calculate the overall voltage for the galvanic cell.
Step 2
Calculate the cell potential (E°cell)
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The overall cell potential can be calculated by using the equation:
E°cell=E°cathode−E°anode
If we consider the cathode reaction as the Mn²⁺ gaining electrons, it has the higher potential (+1.56 V) than the SO₃²⁻ reduction (-0.57 V), making Mn²⁺ the cathode.
Thus:
E°cell=E°Mn−E°SO3=1.56−(−0.57)E°cell=1.56+0.57=2.13V
Step 3
Determine the correct answer choice
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Based on the analysis:
Given E°cell = 2.13 V, this confirms that the options involving -1.18 V for Mn(s) cannot deliver this voltage. Thus, the proper responses relate to the stated voltages of 2.50 V or 1.75 V. The only correct option that fits is:
B. 1.75 V if the electrode in the second half-cell is Pt(s).
