1896 United States presidential election in Nevada

1896 United States presidential election in Nevada

← 1892 November 3, 1896 1900 →
 
Nominee William Jennings Bryan William McKinley
Party Silver Republican
Alliance Democratic
Populist
Home state Nebraska Ohio
Running mate Arthur Sewall (Democratic/Silver)
Thomas E. Watson (Populist)
Garret Hobart
Electoral vote 3 0
Popular vote 8,376 1,938
Percentage 81.21% 18.79%

County Results
Bryan
  60-70%
  70-80%
  80-90%
  90-100%


President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

The 1896 United States presidential election in Nevada took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. State voters chose three electors to the Electoral College, which selected the president and vice president.

Nevada was won by the Silver and Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate Arthur Sewall of Maine. They defeated the Republican nominees, former Governor of Ohio William McKinley and his running mate Garret Hobart of New Jersey. Bryan won the state by a margin of 62.42%, which to date remains the strongest of any presidential nominee in the history of Nevada.

With 81.21% of the popular vote, Nevada would prove to be Bryan's fifth strongest state in the 1896 presidential election only after Mississippi, South Carolina, Colorado and neighboring Utah.[1]

Bryan would later defeat McKinley in the state four years later and would also later defeat William Howard Taft in the state in 1908. This election was the first time ever that the presidential candidate that won Nevada carried all counties in Nevada. This was also the first time that Ormsby County backed a Democratic candidate.

  1. ^ "1896 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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