1847 Vermont gubernatorial election

The 1847 Vermont gubernatorial election took place on September 7, 1847, and resulted in the reelection of Whig Party candidate Horace Eaton to another one-year term as governor, his second.

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1847 Vermont gubernatorial election

← 1846 September 7, 1847 (1847-09-07) 1848 â†’

Nominee Horace Eaton Paul Dillingham Lawrence Brainerd Party Whig Democratic Liberty Electoral vote 125 85 19 Popular vote 22,455 18,601 6,926 Percentage 46.68% 38.67% 14.39%

County results Eaton:      40–50%      50–60%      60–70% Dillingham:      40–50%

Governor before election Horace Eaton Whig Elected Governor Horace Eaton Whig

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In the election for governor, Eaton, who had succeeded from lieutenant governor in 1846, was elected to his second one-year term, defeating Democrat Paul Dillingham, Liberty candidate Lawrence Brainerd, and Know Nothing Reuben C. Benton.^[1]

In the lieutenant governor's election, Whig Leonard Sargeant was elected to a second one-year term with 46.5% of the vote to Democrat Charles K. Field's 39.0%, and the Liberty Party Jacob Scott's 14.3%, with a handful of votes cast for write-in candidates.^[2]

For treasurer, Whig George Howes received 46.4%, Democratic nominee received Jeremiah T. Marston 39.1%, and Liberty nominee Zenas Wood had 14.3%, with a few ballots cast for write-ins.^[3]

Because none of the candidates received the popular vote majority required by the Vermont Constitution, the contests were formally decided by the Vermont General Assembly.^[4] In the October 14 vote, 229 representatives and senators took part, so 115 votes were required to win.^[4] Eaton, Sargeant, and Howes were all elected on the first ballot.^[4] Eaton received 125 votes, Sargeant 124, and Howes 124.^[4]