Answer:
1219.5 kj/mol
Explanation:
To reach this result, you must use the formula:
ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).
The BE values are:
BE C = C: 839 kj / mol
BE C-H: 413 Kj / mol
BE O = O: 495 kj / mol
BE C = O = 799 Kj / mol
BE O-H = 463 kj / mol
Now you must replace the values in the above equation, the result of which will be:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol
The question is missing. Here is the complete question.
Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of
?
(a) 
(b) 
(c)
(d) 
Answer: (d) 
Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.
Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.
For the cell notation: 
Aluminum's half-equation is oxidation:

For Lead, half-equation is reduction:

Multiply first half-equation for 2 and second half-equation by 3:


Adding them:

The balanced redox reaction with cell notation
is

Answer:
C is the element thats has been oxidized.
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)
This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.
In the oxalic acid, this are the oxidation states:
H: +1
C: +3
O: -2
In the product side, in CO₂ the oxidation states are:
C: +4
O: -2
Carbon from the oxalate has increased the oxidation state, so it has been oxidized.