Home
Class 11
CHEMISTRY
Number of electron involved in the reduc...

Number of electron involved in the reduction of `Cr_(2)O_(7)^(2-)` ion in acidic solution to `Cr^(3+)` is:

Promotional Banner

Similar Questions

Explore conceptually related problems

Number of electron involved in the reduction of Cr_(2)O_(7)^(2-) ion in acidic solution of Cr^(3+) is

Number of electrons involved in the reduction of Cr_(2)O_(7)^(2-) ion in acidic solution to Cr^(3+) is

Number of electron involved in the reduction of Cr_(2)O_(7)^(2-) ion in acidic solution of Cr^(3+) is

Number of electrons involved in the reduction of Cr_2O_7^(2-) ion in acidic solution to Cr^(3+) is:

The number of electrons involved in the half reaction Cr_(2)O_(7)^(2-)rarr2Cr^(3+) is

The charge required for the reduction of 1 mol Cr_(2)O_(7)^(2-) ions to Cr^(3+) is

The charge required for the reduction of 1 mol Cr_(2)O_(7)^(2-) ions to Cr^(3+) is

The hybridisation of Cr in Cr_2O_7^(2-) ion is