An iron strip (in excess) is dipped into 100 mL of 0.200 M CuSO4 and left until the reaction is complete. Given E deg(Fe^2+/Fe) = -0.44, E deg(Fe^3+/Fe^2+) = 0.77, E deg(Cu^2+/Cu) = 0.34 V, decide the oxidation state of the iron product and find the net change in mass of the strip (in g, two decimals; a gain is positive). Fe = 56, Cu = 63.5.

JEE Advanced levelNumerical answerDifficulty 3 of 3ec044

An iron strip (in excess) is dipped into $100 mL$ of $0.200 M$ $CuSO X_{4}$ and left until the reaction is complete. Given $E^{\circ} (Fe X^{2 +} / Fe) = - 0.44$ , $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77$ , $E^{\circ} (Cu X^{2 +} / Cu) = 0.34 V$ , decide the oxidation state of the iron product and find the net change in mass of the strip (in g, two decimals; a gain is positive). Fe = 56, Cu = 63.5.

Numerical answer type, answer in g, tolerance 0.01.

Cu X^{2 +}

(

0.34 V

) can oxidise Fe to

Fe X^{2 +}

(

E_{cell}^{\circ} = 0.34 + 0.44 = 0.78 V

) but not

Fe X^{2 +}

Fe X^{3 +}

(

0.34 - 0.77 < 0

), so the product is

Fe X^{2 +}

Fe + Cu X^{2 +} Fe X^{2 +} + Cu

. Moles of

Cu X^{2 +} = 0.100 \times 0.200 = 0.0200

. Mass of Cu deposited

= 0.0200 \times 63.5 = 1.27 g

; mass of Fe dissolved

= 0.0200 \times 56 = 1.12 g

. Net change

= 1.27 - 1.12 = + 0.15 g