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pychu [463]
2 years ago
13

A voltaic cell is based on the following two half-reactions: ni 2(aq) 2e− → ni(s) e° = −0. 25 v cr 3(aq) 3e– → cr(s) e° = –0. 74

v sketch the cell and then select the correct statement about it
Chemistry
1 answer:
emmasim [6.3K]2 years ago
3 0

The net cell reaction is given as 3 Ni²⁺ (aq) + 2 Cr (s) = 2 Cr³⁺ (aq) + 3 Ni (s), and the cell potential is 0.49 V.

<h3>What is a half-cell reaction?</h3>

A half cell reaction is given as the oxidation or the reduction reaction that takes place at each of the anode or the cathode, forming the net redox reaction.

The cell potential and the half-reactions are:

Ni²⁺ + 2e⁻ = Ni (s)     E° = -0.25 V

Cr³⁺ + 3e⁻ = Cr (s)     E° = -0.74 V

The anodic reaction is given as:

Cr (s) = Cr³⁺ + 3e⁻     E° = 0.74 V

The cathodic reaction is given as:

Ni²⁺ + 2e⁻ = Ni (s)     E° = -0.25 V

Thus, the net reaction of the cell will be:

3 Ni²⁺ (aq) + 2 Cr (s) = 2 Cr³⁺ (aq) + 3 Ni (s)

The cell potential can be calculated as:

E° = E (cathode) + E (anode)

E° = 0.74 + (-0.25) V

E° = 0.49 V

Thus, the net cell reaction is given as 3 Ni²⁺ (aq) + 2 Cr (s) = 2 Cr³⁺ (aq) + 3 Ni (s), and the cell potential is 0.49 V.

Learn more about cell potential, here:

brainly.com/question/1313684

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Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume
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1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

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