For reaction :
`Cr_(2)O_(7)^(-2)+14H^(+)rarr2Cr^(+3)+7H_(2)O`, How many `e^(-)s` are required

Consider the reaction : Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol e of Cr_(2)O_(7)^(2-) ?

Which substance is serving as a reducing agent in the following reaction ? Cr_(2) O_(7)^(2-) + 14H^(+) + 3Ni rarr 2Cr^(3+) + 7H_(2)O + 3Ni^(2+)

In the following reaction 2I- + Cr_(2)O_(7)^(2-)+14H^(+) rarr I_(2)+2Cl^(3)+7H_(2)O Unbalanced parts are

The electrode with reaction :Cr_(2)O_(7)^(2-)(aq)+14H^(o+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_(2)O can be represented as

The electrode with reaction :Cr_(2)O_(7)^(2-)(aq)+14H^(o+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_(2)O can be represented as

In the following reaction Cr_(2) O_(7)^(2-) + 14H^(+) + 6I^(-) rarr 2Cr^(3+) + 7H_(2)O + 3I_(2) Which element gets reduced ?

Consider the reaction : Cr_(2)O_(7)^(2-)+14H^(o+)+6e^(-) rarr 2Cr^(+3)+8H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol e of Cr_(2)O_(7)^(2-) ?

Consider the reaction, Cr_(2) O_(7) ^(2-) + 14 H^(+) + 6e^(-) rarr 2Cr^(3+) + 7 H_(2) O What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr_(2)O_(7)^(2-) ?