The reaction involves the oxidation of `Sn^(2+)to Sn^(4+)` ion the skeleton equation for the reaction is :
`Cr_(2)O_(7)^(2-)+Sn^(2+)+H^(+)rarrCr^(3+)+Snb^(4+)+H_(2)O`
Balance the equation by ion electron method.

Explanation

Separtion of the equation in two half reactions
(i) Write the O.N of the atoms involved in the equation:
`(overset(+6)Cr_(2)overset(-2)O_(7))^(2-)+(overset(+2)Sn)^(2+)+(overset(+1)H)rarr(overset(+3)Cr)^(3+)+(overset(+4)Sn)^(4+)+overset(+1)H_(2)overset(-2)O`
(ii) Identify the atoms which undergo change in O.N .
`(overset(+6)Cr_(2)overset(-2)O_(7))^(2-)+(overset(+2)Sn)^(2+)+H^(+)rarr(overset(+3)Cr)^(3+)+(overset(+4)Sn)^(4+)+H_(2)O`
(iii) Find out the speicees involved in the oxidation and reduction half reactions

Oxidation half reaction: `Sn^(2+)rarrSn^(4+)`
Reduction half reaction: `Cr_(2)O_(7)rarrCr^(3+)`
Step II.Balancing the oxidation half reaction
The oxidation half reaction is : `Sn^(2)rarrSn^(4+)`
(i) as the increase in O.N as a result of oxidation is 2, add two `e^(-)` on the product side to balance the change in O.N
`Sn^(2)rarrSn^(4+)+2e^(-)`
(ii) Since the charge is already, therefore the equation is blanced.
`Sn^(2+)rarrSn^(4+)+2e^(-)`
Step II .Balancing of reduction half reaction
The reduction half reaction is : `(overset(+6)Cr_(2)O_(7))^(2-)rarr(overset(+3)Cr)^(3+)`
(i) The decrease in O.N per Cr atom is 3 and the total decrease in O.N for two Cr atoms is 6. Therefore , add `6e^(-)` on the reactant side
`(Cr_(2)O_(7))^(2-)+6e^(-)rarrCr^(3+)`
(ii) Balance Cr atoms on both sides of the equation
`(Cr_(2)O_(7))^(2-)+6e^(-)rarr2Cr^(3+)`
(iii) In order to balance O atoms add seven `H_(2)O` molecules on the product side and then to balance H atoms add 14 `H^(+)` on the reactant side.
`(Cr_(2)O_(7))^(2-)+6e^(-)14H^(+)rarr2Cr^3+7H_(2)O`
Step IV Adding the two half reaction:
In order to tequation the electrons, multiply eqn. (i) by 3 and add to equ(ii)
`{:(" "Sn^(2+)rarrSn^(4+)+de^(-)"]"xx3),(Cr_(2)O_(7)^(2-)+6e^(-)+14H^(+)rarr2Cr^(3)+7H_(2)O))/(3Sn^(2+)+(Cr_(2)O_(7))^(2-)+14Hrarr3Sn^(4+)+2Cr^(3+)+7H_(2)O)`