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solniwko [45]
3 years ago
14

An iron bar at 325 K is placed in a sample of water. The iron bar gains energy from the water if the temperature of the water is

Chemistry
1 answer:
noname [10]3 years ago
3 0
So what you need to do is to have all your answers in the same unit as the one in the question which is K then look for the one that is greater than that in the question because only then will it absorb heat from the water.

temp. of rod:  325K
(1)  65 K
(2) 45 K
(3) 65° C  =  (65 + 273) K = 338 K
(4) 45° C  =  (45 + 273) K = 318 K

⇒ option 3 is the answer
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A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:
seropon [69]

<u>Answer:</u> The cell voltage of the given reaction is 1.86 V

<u>Explanation:</u>

The given chemical equation follows:

3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)

<u>Oxidation half reaction:</u> Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V       ( × 2)

<u>Reduction half reaction:</u> Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V       ( × 3)

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}=0.16-(-1.66)=1.82V

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

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}

where,

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

E^o_{cell} = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = 42^oC=[42+273]K=315K

F = Faraday's constant = 96500

[Al^{3+}]=1.63M

[Cu^{2+}]=3.43M

Putting values in above equation, we get:

E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V

Hence, the cell voltage of the given reaction is 1.86 V

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