All categories

3 years ago

Answer and explanation please! This is help for a upcoming quiz!

Chemistry

1 answer:

Norma-Jean [14]3 years ago

3 0

I’m so sorry this is confusing

You might be interested in

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

Answer: The potential of the given cell is 1.551 V

Explanation:

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

Reduction half reaction: $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

Net cell reaction: $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

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:

$E^o_{cell}=0.799-(-0.762)=1.561V$

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

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago

Spell out the full name of the compound.

ELEN [110]

Idk but I’ll try to answer

3 0

3 years ago

which of the following makes ice cores so useful in determining the composition of the atmosphere in the past?

iVinArrow [24]

Air bubbles, i just had this question on a test.

3 0

3 years ago

Read 2 more answers

A car tire has a pressure of 2.38 atm at 15.2 c. If the pressure inside reached 4.08 atm, the tire will explode. How hot would t

stich3 [128]

The correct answer is 221.06 °C hot.

If P₁ is the pressure at T₁ and P₂ is the pressure at T₂ then,

P₁/T₁ = P₂/T₂

It is given that P₁ = 2.38 atm

T₁ = 15.2 degree C = 273 + 15.2 = 288.2 K

P₂ = 4.08 atm

T₂ = x

Thus, 2.38 / 288.2 = 4.08 / x

x = (4.08 × 288.2) / 2.38

x = 494.06 K

x = 494.06 - 273 °C = 221.06 °C

Therefore, the tire would get 221.06 °C hot.

8 0

3 years ago

Reactant P contains 50 J of energy, and reactant Q contains 35.3 J of energy. The reactants combine to form product PQ, which co

Drupady [299]

85.3 i hope its add it all than minis

5 0

4 years ago