**Contents**show

## How much charge is required for the reduction of 1 mole of mno4 to Mn2+?

**93×105C**.

## How many coulombs are required for reduction of 1 mole of mno4 to 2 mno4?

**62×105C**.

## How many moles of electrons are required to a reduce 1 mol of mno4 to mno2?

Hence **5 moles** of electrons are required for 1 mole of MnO−4 .

## How much charge is required for the reduction of 1 mole of Cu 2 ion to see you?

How much charge is required for the reduction of 1 mole of Cu^(2+) to Cu. =2×96500=**193000C**.

## How much charge is required for the reduction of 1 mole?

Therefore, **5 mol of electrons** are required to reduce 1 mol of MnO-4 ions).

## What amount of electric charge is required for the reduction of 1 mole of cr2o72 to cr3+?

**9650C**.

## What is the charge of 1 mole of mno4?

Complete step by step solution:

We can see that here, $MnO_{4}^{ – }$ is 1 mole and there is 5 mole of the electron. – Here, we can see that in $MnO_{4}^{ – }$ the oxidation state is **+7** and in $M{{n}^{2+}}$ oxidation state is +2, so there is a change in oxidation number taking place.

## How many Faraday of electricity is required to reduce 1 mole of mno4 minus ions to Mn 2 ions?

How many faradays are required to reduce one moleof MnO_(4)^(-) to Mn^(2+)? Number of faradays is equal to the change in oxidation state. Since oxidation state of Mn decreases by 5 units (from +7 to +2), **five faradays** is needed.

## How much charge is required for the reduction of mno4 to mno2?

**1F**. **3F**.

## How much electricity is required in column for the oxidation of 1 mole of h2o to o2?

<br> The charge Q on n mol of electrons is given by Q=nF <br> Thus, quantity of electricity required for oxidation of 1 mol of `H_(2)O “to” O_(2)`, <br> **`Q=2 mol xx96500 “C mol”^(-1)=193000 C =1.93xx10^(5)C`**.

## How many coulombs are required for the reduction of 1 mole of cu2+ to CU?

∴ For the reduction of 1 mole ofCu+ ions to Cu, charge required is **2F**.

## How many electrons does mno4 have?

Total valence electrons pairs

Total electron pairs are determined by dividing the number total valence electrons by two. For, MnO_{4}^{–} ion, total pairs of electrons are **16**.