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Chemistry
Chemistry
Problem
Dilute sulfuric acid is electrolysed using inert electrodes. What are the ionic half-equations for the reactions that take place at each electrode?
positive electrode
negative electrode
A
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
B
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
4
OH
−
+
4
H
+
→
4
H
2
O
4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O}
4
OH
−
+
4
H
+
→
4
H
2
O
C
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
D
4
OH
−
+
4
H
+
→
4
H
2
O
4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O}
4
OH
−
+
4
H
+
→
4
H
2
O
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
A) Positive:
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
; Negative:
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
B) Positive:
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
; Negative:
4
OH
−
+
4
H
+
→
4
H
2
O
4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O}
4
OH
−
+
4
H
+
→
4
H
2
O
C) Positive:
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-
4
OH
−
→
2
H
2
O
+
O
2
+
4
e
−
; Negative:
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
D) Positive:
4
OH
−
+
4
H
+
→
4
H
2
O
4\text{OH}^- + 4\text{H}^+ \rightarrow 4\text{H}_2\text{O}
4
OH
−
+
4
H
+
→
4
H
2
O
; Negative:
2
H
+
+
2
e
−
→
H
2
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
2
H
+
+
2
e
−
→
H
2
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