Definition: This is the male reproductive cell.

Calculate Ecell for the following electrochemical cell at 25 ºCPt (s) | H2 (g, 1.00 atm) | H+ (aq, 1.00 M) || Sn2+ (aq, 0.350 M)

katrin2010 [14]

Answer: The electrode potential of the cell is 0.093 V

Explanation:

The given chemical cell follows:

$Pt(s)|H_2(g,1atm)|H^+(aq,1.00M)||Sn^{4+}(aq,0.020M)|Sn^{2+}(aq.,0.350M)|Pt(s)$

Oxidation half reaction: $H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V$

Reduction half reaction: $Sn^{4+}(aq,0.020M)+2e^-\rightarrow Sn^{2+(aq.,0.350M);E^o_{Sn^{4+}/Sn^{2+}}=0.13V$

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.13-(0.0)=0.13V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2[Sn^{2+}]}{[Sn^{4+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +0.13 V

n = number of electrons exchanged = 2

$[H^{+}]=1.00M$

$[Sn^{2+}]=0.350M$

$[Sn^{4+}]=0.020M$

Putting values in above equation, we get:

$E_{cell}=0.13-\frac{0.059}{2}\times \log(\frac{(1.0)^2\times 0.350}{0.020})$

$E_{cell}=0.093V$

Hence, the electrode potential of the cell is 0.093 V

3 0

4 years ago

What direction would this plate move? (the plate is gray in color)

strojnjashka [21]

Answer:

from south to north thats what it look like

Explanation:

7 0

3 years ago

Which is the net result of the proton-proton chain? 6 protons = 2 heliums + 3 positrons + 3 neutrinos + gamma rays 4 protons = 1

saveliy_v [14]

Answer: 5.125 neutrinos

Explanation:

8 0

3 years ago

The rate constant for a particular zero-order reaction is 0.075 M s-1. If the initial concentration of reactant is 0.537 M it ta

Phoenix [80]

Answer : It takes time for the concentration to decrease to 0.100 M is, 22.4 s

Explanation :

Formula used to calculate the rate constant for zero order reaction.

The expression used is:

$\ln [A]=-kt+\ln [A_o]$

where,

$[A_o]$ = initial concentration = 0.537 M

$[A]$ = final concentration = 0.100 M

t = time = ?

k = rate constant = 0.075 M/s

Now put all the given values in the above expression, we get:

$\ln (0.100)=-0.075\times t+\ln (0.537)$

$t=22.4s$

Therefore, it takes time for the concentration to decrease to 0.100 M is, 22.4 s

6 0

3 years ago

If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?

kodGreya [7K]

Answer:

$[Cu^{2+}]=0.041 M$

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

$[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}$

$[Cu^{2+}]=0.041 M$

Best regards!

6 0

3 years ago