Question

Calculate the potential when 25.00 mL of 0.0750 M Cut is titrated with 12.50 mL of 0.150 M Fe3+ in the presence of 1 M HCl, using Pt and saturated calomel electrodes to find the end point. Fe3 Fe E+0732 v Fe3t +e Fe2+ E0.732v Cu2+ +eCut E +0.161 V a +0.401 V 0 b. +0.206 V +1.124v d. +0.157 V

Answer 1

$Fe^{3+}+Cu^{+}\rightarrow Fe^{2+}+Cu^{2+}$

From the data given , here we can see , the addition of 12.5 ml of Fe³⁺ solution on 25 ml Cu⁺ solution gives us the equivalence point .

The equation is =

$(25 ml)\times (0.0750M)=(12.5 ml)\times (0.150M)$

The cell potential under equivalence condition is given by the equation -

$E^{0}=(n_{1}E_{1}+n_{2}E_{2})\div (n_{1}+n_{2})$

Here n_{1 i}s the no. of electron transfer by iron couple , that is , 1

n₂ is the no. of electron transfer by copper couple , that is , 1

E₁ is the redox potential for iron couple , that is +0.732 V

E₂ is the redox potential for copper couple , that is +0.161 V

Thus , $E^{0}=0.4465 V$

Now overall cell EMF is the rest EMF calculating omitting the Saturated Calomel Electrode's (SCE) potential , that is at STP +0.2444V.

Thus , overall cell EMF--

$E=E^{0}-E_{SCE}=(0.4465-0.2444)=0.2021V$

Our calculated answer is closest to option b (0.206 V)

So , final answer is Option B - 0.206 V .