Answer:
Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻
Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)
E° = 1.60 V
Explanation:
Let's consider the reaction taking place in a galvanic cell.
2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)
The corresponding half-reactions are:
Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻ E°red = - 0.83 V
Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq) E°red = 0.77 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V
Answer : The amount of heat absorbed are, 258485.5 J
Solution :
Formula used :
where,
Q = heat gained = ?
m = mass of water = 650 g
c = specific heat of water = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get the final temperature of copper.
Therefore, the amount of heat absorbed are, 258485.5 J
Answer:
Me
Explanation:
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