Question

please help the standrad reduction provided down

What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cu2+ concentration is 9.24x10 + M and the Cr3+ concentration is 1.47 M? 3Cu2+ (aq) + 2Cr(s) Answer: 3Cu(s) + 2Cr +(aq) v The cell reaction as written above is spontaneous for the concentrations given:

Answer 1

1) Cu²⁺ ( aq ) + 2e^- $\rightarrow$ Cu ( s )   E^o = 0.337 volts

Cr³⁺ ( aq) + 3e^- $\rightarrow$ Cr ( s ) E⁰ = —0.74 volts

E^o_cell = E^o ( Cu²⁺ / Cu ) — E^o ( Cr³⁺ / Cr ) = 0.337 volt — ( —0.74 volt ) = 1.077 volt

using Nernst equation

E_cell = E⁰_cell —( RT / nF ) ln {[ Cr³⁺]² / [ Cu²⁺ ]³ }
   = 1.077 volt —( 0.059 / 6 ) log [(1.47 )² / ( 9.24 x 10^–4 )³ ]
( since 2.303RT / F = 0.059 and number of electrons transferred = 6)

= 1.077 volt — 0.01 volt = 1.087 volt ( spontaneous cell reaction)

( b) Hg²⁺ ( aq ) + 2e^- $\rightarrow$ Hg ( s ). E^o = 0.855 volt

Mn²⁺ ( aq ) + 2e^- $\rightarrow$ Mn ( s ) E^o = —1.18 volt

E^o_{cell =} E^o ( Hg²⁺ / Hg ) — E^o ( Mn²⁺ / Mn )

= 0.855 volt — ( —1.18 volt ) = 2.035 volt

E_cell = E^o_cell — (0.059 / 2 ) log {[ Mn²⁺ ] / [ Hg²⁺ ] }

= 2.035 volt — ( 0.059 / 2 ) log ( 3.93 x 10^–4 M / 1.19 M )

= 2.137 volt ( spontaneous cell reaction)