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

Elections in Nevada
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None of These Candidates  Nevada portal
v t e

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 in term remains as of 2023 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.